Respuesta :
a). 1.22%
b). 2%
c). 3.72%
d). 5.94%
Further explanation
Given:
Ka for formic acid, HCOOH, is [tex]1.80 \times 10^{-4}[/tex]
Question:
Calculate the percent ionization (α) of formic acid solutions having the following concentrations (M).
The Process:
Remember that,
[tex]\boxed{ \ [H^+] = \alpha M \ }[/tex] and [tex]\boxed{ \ [H^+] = \sqrt{K_a M} \ }[/tex]
The two equations are rearranged to connect α, Ka, and M directly.
[tex]\boxed{ \ \alpha M = \sqrt{K_a M} \ }[/tex]
[tex]\boxed{ \ (\alpha M)^2 = (\sqrt{K_a M})^2 \ }[/tex]
[tex]\boxed{ \ \alpha^2 M^2 = K_a M \ }[/tex]
[tex]\boxed{ \ \alpha^2 = \frac{K_a M}{M^2} \ }[/tex]
[tex]\boxed{\boxed{ \ \alpha = \sqrt{\frac{K_a}{M}} \ }}[/tex]
Let us calculate the percent ionization from each concentration below.
a). 1.20 M
[tex]\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{1.20}} \ }[/tex]
[tex]\alpha \approx 0.0122[/tex]
Thus, the percent of ionization for 1.20 M HCOOH solution is 1.22%.
_ _ _ _ _ _ _
b). 0.450 M
[tex]\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{0.450}} \ }[/tex]
[tex]\alpha \approx 0.02[/tex]
Thus, the percent of ionization for 0.450 M HCOOH solution is 2%.
_ _ _ _ _ _ _
c). 0.130 M
[tex]\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{0.130}} \ }[/tex]
[tex]\alpha \approx 0.0372[/tex]
Thus, the percent of ionization for 0.130 M HCOOH solution is 3.72%
_ _ _ _ _ _ _
d). 5.10 x 10⁻² M
[tex]\boxed{ \ \alpha = \sqrt{\frac{1.80 \times 10^{-4}}{5.10 \times 10^{-2}}} \ }[/tex]
[tex]\alpha \approx 0.0594[/tex]
Thus, the percent of ionization for 5.10 x 10⁻² M HCOOH solution is 5.94%.
_ _ _ _ _ _ _ _ _ _
Notes
Just a reminder, Ka and α are characteristic of weak acid solutions.
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The percent ionization of formic acid solutions having the concentrations of 1.20, 0.450, 0.130, and 5.10x10⁻² M is 1.22, 1.98, 3.61, and 5.69%, respectively.
The reaction of dissociation of formic acid is the following.
CHOOH + H₂O ⇄ CHOO⁻ + H₃O⁺
In the equilibrium we have:
CHOOH + H₂O ⇄ CHOO⁻ + H₃O⁺ (1)
Ci - x x x
[tex] K_{c} = \frac{[CHOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x*x}{(C_{i} - x)} [/tex] (2)
Where:
Kc: is the equilibrium constant of formic acid = 1.8x10⁻⁴
Ci: is the initial concentration of formic acid
The percent ionization of formic acid can be calculated with the following equation:
[tex] \%_{I} = \frac{[H_{3}O^{+}]}{C_{i}} \times 100 [/tex] (3)
With the above information, we can find the percent ionization of formic acid of each different concentration.
a) Ci = 1.20 M
We can find the H₃O⁺ concentration with equation (2)
[tex] 1.8 \cdot 10^{-4} = \frac{x^{2}}{(C_{i} - x)} [/tex]
[tex] 1.8 \cdot 10^{-4}(1.20 - x) - x^{2} = 0 [/tex]
After solving the above quadratic equation for x, we have:
[tex] x_{1} = 0.0146 M = [H_{3}O^{+}] [/tex]
[tex] x_{2} = -0.0146 M = [H_{3}O^{+}] [/tex]
Taking the positive value (x₁) we can find the percent ionization (eq 3)
[tex] \%_{I} = \frac{[H_{3}O^{+}]}{C_{i}} \times 100 = \frac{0.0146}{1.20} \times 100 = 1.22 \% [/tex]
Hence, the percent ionization of formic acid is 1.22% when the initial concentration is 1.20 M.
b) Ci = 0.450 M
From equation (2) we have:
[tex] 1.8 \cdot 10^{-4}(0.450 - x) - x^{2} = 0 [/tex]
After solving the above quadratic equation for x, we have:
[tex] x_{1} = 0.0089 M = [H_{3}O^{+}] [/tex]
[tex] x_{2} = -0.0090 M = [H_{3}O^{+}] [/tex]
With the positive value we can find the percent ionization (eq 3)
[tex] \%_{I} = \frac{0.0089}{0.450} \times 100 = 1.98 \% [/tex]
Therefore, the percent ionization of formic acid is 1.98% when the initial concentration is 0.450 M.
c) Ci = 0.130 M
From eq (2) we have:
[tex] 1.8 \cdot 10^{-4}(0.130 - x) - x^{2} = 0 [/tex]
After solving the above quadratic equation for x:
[tex] x_{1} = 0.0047 M = [H_{3}O^{+}] [/tex]
[tex] x_{2} = -0.0049 M = [H_{3}O^{+}] [/tex]
The percent ionization is (eq 3):
[tex] \%_{I} = \frac{0.0047}{0.130} \times 100 = 3.61 \% [/tex]
Then, the percent ionization of formic acid is 3.61% when the initial concentration is 0.130 M.
d) Ci = 5.10x10⁻² M
The H₃O⁺ concentration is:
[tex] 1.8 \cdot 10^{-4}(5.10 \cdot 10^{-2} - x) - x^{2} = 0 [/tex]
[tex] x_{1} = 0.0029 M = [H_{3}O^{+}] [/tex]
[tex] x_{2} = -0.0031 M = [H_{3}O^{+}] [/tex]
So, the percent ionization is (eq 3):
[tex] \%_{I} = \frac{0.0029}{5.10 \cdot 10^{-2}} \times 100 = 5.69 \% [/tex]
The percent ionization of formic acid is 5.69% when the initial concentration is 5.10x10⁻² M.
Therefore, the percent ionization of formic acid solutions for the concentrations 1.20, 0.450, 0.130, and 5.10x10⁻² M is 1.22, 1.98, 3.61, and 5.69%, respectively.
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